Giải thích các bước giải:

1.Xét $\Delta AID,\Delta AIH$ có:

Chung $AI$

$\widehat{AID}=\widehat{AIH}(=90^o)$

$ID=IH$

$\to\Delta AID=\Delta AIH(c.g.c)$

b.Từ câu a

$\to AD=AH,\widehat{DAI}=\widehat{IAH}\to\widehat{DAB}=\widehat{BAH}$

Xét $\Delta ADB,\Delta AHB$ có:
Chung $AB$

$\widehat{DAB}=\widehat{BAH}$

$AD=AH$

$\to\Delta ADB=\Delta AHB(c.g.c)$

$\to\widehat{ADB}=\widehat{AHB}=90^o$

$\to BD\perp AD$

c.Tương tự câu b chứng minh được

$\Delta AEC=\Delta AHC(g.c.g), CE\perp AE$

$\to \widehat{HAC}=\widehat{CAE}$

$\to \widehat{HAE}=2\widehat{HAC}$

Mặt khác từ câu $b\to\widehat{DAB}=\widehat{BAH}\to \widehat{DAH}=2\widehat{BAH}$

$\to \widehat{DAE}=\widehat{DAH}+\widehat{HAE}=2\widehat{BAH}+2\widehat{HAC}=2\widehat{BAC}=180^o$

$\to D,A,E$ thẳng hàng

Mà $BD\perp AD, CE\perp AE\to BD\perp DE, CE\perp DE$

$\to BD//CE$